The present population of a town is 176400. If the rate of growth in its population is 5% per annum, find:

(i) its population 2 years hence,

(ii) its population one year ago.

(i) Given,

P = 176400

R = 5% p.a.

n = 2 years

By formula,

Population after n years = $P \times \Big(1 + \dfrac{r}{100}\Big)^n$

Substituting the values in formula,

$\text{Population after 2 years}=176400 \times \Big(1 + \dfrac{5}{100}\Big)^2 \\[1em] =176400 \times \Big(\dfrac{100 + 5}{100}\Big)^2 \\[1em] =176400 \times \Big(\dfrac{105}{100}\Big)^2 \\[1em] =176400 \times \Big(\dfrac{21}{20}\Big)^2 \\[1em] =176400 \times \dfrac{441}{400} \\[1em] =194481$

Hence, population of the town after 2 years = 194481.

(ii) Given,

P = 176400

R = 5% p.a.

n = 1 year

By formula,

Population before n years = $\dfrac{P}{\Big(1 + \dfrac{r}{100}\Big)^n}$

Substituting the values in formula,

$\text{Population one year ago} =\dfrac{176400} {\Big(1 + \dfrac{5}{100}\Big)} \\[1em] = \dfrac{176400} {\Big(\dfrac{105}{100}\Big)} \\[1em] = \dfrac{176400} {\dfrac{21}{20}} \\[1em] = \dfrac{176400 \times 20} {21} \\[1em] = 168000$

Hence, population of the town before 1 year = 168000.